A bob of mass 10 M is suspended through an inextensible string of length ℓ. When the bob is at rest at the equilibrium position, two particles of mass m each moving with velocity u making an angle 60° with the string strike and get simultaneously attached to the bob. What is the value of impulsive tension (ℓ) in the string during the impact?
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During the impact, two particles of mass m each strike the bob symmetrically with velocity u at 60° to the string. The horizontal components of their velocities cancel out due to symmetry, so no net horizontal impulse is imparted. Only the vertical components contribute to the impulsive tension. Each particle has a vertical velocity component of u sin60° = (√3/2)u. The total vertical impulse from both particles is 2 * m * (√3/2)u = √3 m u. However, the question asks for impulsive tension, and since the vertical impulse is balanced by the tension impulse, the value is √3 m u, but reviewing the options, none match exactly. The correct approach shows the impulsive tension T satisfies T Δt = change in vertical momentum. The vertical momentum change is 2m(u cos60°) = 2m(u/2) = m u. Thus, impulsive tension is m u.
Final answer: mu